poj 3261 Milk Patterns （后缀数组）

Milk Patterns

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 14691 Accepted: 6535Case Time Limit: 2000MS

Description

Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K times.

Input

Line 1: Two space-separated integers: N and K 
Lines 2..N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at least K times

Sample Input

8 212323231

Sample Output

4

Source

USACO 2006 December Gold

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题目大意：给出一个字符串，求出现至少k次的的最长重复子串，子串可重叠。

题解：后缀数组

和上一道题类似，也要用后缀数组+二分。二分答案，将height分组，然后判断每组中的个数是否>=k。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#define N 20003using namespace std;int n,m,k,len,a[N],sa[N],rank[N],height[N],xx[N],yy[N],*x,*y,p;int b[1000003];int cmp(int i,int j,int l){return y[i]==y[j]&&(i+l>len?-1:y[i+l])==(j+l>len?-1:y[j+l]);}void get_SA(){x=xx; y=yy; m=1000000;for (int i=1;i<=n;i++) b[x[i]=a[i]]++;for (int i=1;i<=m;i++) b[i]+=b[i-1];for (int i=n;i>=1;i--) sa[b[x[i]]--]=i;for (int k=1;k<=n;k<<=1) {p=0;for (int i=n-k+1;i<=n;i++) y[++p]=i;for (int i=1;i<=n;i++) if (sa[i]>k) y[++p]=sa[i]-k;for (int i=1;i<=m;i++) b[i]=0;for (int i=1;i<=n;i++) b[x[y[i]]]++;for (int i=1;i<=m;i++) b[i]+=b[i-1];for (int i=n;i>=1;i--) sa[b[x[y[i]]]--]=y[i];swap(x,y); p=2; x[sa[1]]=1;for (int i=2;i<=n;i++) x[sa[i]]=cmp(sa[i-1],sa[i],k)?p-1:p++;if (p>len) break;m=p+1; }p=0;for (int i=1;i<=n;i++) rank[sa[i]]=i;for (int i=1;i<=n;i++) {if (rank[i]==1) continue;int j=sa[rank[i]-1];while (i+p<=n&&j+p<=n&&a[i+p]==a[j+p]) p++;height[rank[i]]=p;p=max(0,p-1);}}int pd(int x){int size=1;for (int i=2;i<=n;i++) if (height[i]>=x) size++; else { if (size>=k) return 1; size=1; } if (size>=k) return 1;return 0;}int main(){freopen("a.in","r",stdin);freopen("my.out","w",stdout);scanf("%d%d",&n,&k); len=n;for (int i=1;i<=n;i++) scanf("%d",&a[i]);get_SA();int l=0; int r=n; int ans=0;while (l<=r) {int mid=(l+r)/2;if (pd(mid)) ans=max(ans,mid),l=mid+1;else r=mid-1;}printf("%d\n",ans);}